Which is the correct way to write $450.05 in words on a check?

The parent function f(x)=log^3x has been transformed by reflecting it over the X axis, stretching it vertically by a factor of t

NARA [144]

Answer:

b

Step-by-step explanation:

6 0

3 years ago

Where my math wizzes at? I really need help and fast. A dialation has a center (0, 0, 0). Find the image of the point (-1, 2, 0)

defon

Multiply every coordinate by the scale factor 3

(-1,2,0) ----> (-3,6,0) is the answer

7 0

3 years ago

3(2y-3)=27can u solve this one

navik [9.2K]

3(2y-3)=27
6y-9=27
6y=36
y=6

7 0

3 years ago

At a local university, a sample of 49 evening students was selected in order to determine whether the average age of the evening

pav-90 [236]

Answer:

$z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4$

$p_v =2*P(z>4)=0.0000633$

When we compare the significance level $\alpha=0.1$ we see that $p_v$ so we can reject the null hypothesis at 10% of significance. So the the true mean is difference from 21 at this significance level.

Step-by-step explanation:

Data given and notation

$\bar X=23$ represent the sample mean

$\sigma=3.5$ represent the population standard deviation

$n=49$ sample size

$\mu_o =21$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average age of the evening students is significantly different from 21, the system of hypothesis would be:

Null hypothesis: $\mu = 21$

Alternative hypothesis: $\mu \neq 21$

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z>4)=0.0000633$

Conclusion

When we compare the significance level $\alpha=0.1$ we see that $p_v$ so we can reject the null hypothesis at 10% of significance. So the the true mean is difference from 21 at this significance level.

8 0

3 years ago

A company that produces dog food claims that its bags contain 11 kg of food. From

andriy [413]

By testing the hypothesis we can conclude that the bag does not contain 11 kg of food.

Given mean of 10.6 kg, population standard deviation of 0.6 and sample size of 25.

We are required to find whether the company is right in saying that their bags contain 11 kg of food.

First we have to make hypothesis.

$H_{0}$ :μ≠11

$H_{1}$ :μ=11

We have to use t statistic because the sample size is less than 30.

t=(X-μ)/s/ $\sqrt{n}$

We will use s/ $\sqrt{n}$ =0.6 because we have already given population standard deviation of weights.

t=(11-10.6)/0.6

=0.4/0.6

=0.667

Degree of freedom=n-1

=25-1

=24

T critical at 0.05 with degree of freedom 24=2.0639

T critical at 0.05 with degree of freedom is greater than calculated t so we will accept the null hypothesis.It concludes that the bags donot contain 11 kg of food.

Hence by testing the hypothesis we can conclude that the bag does not contain 11 kg of food.

Learn more about t test at brainly.com/question/6589776

#SPJ1

3 0

2 years ago